Proof. Without restriction assume \(t_0=0\) . Let \(\varphi \)  be a state localized in \(\Delta _0\) , and denote by \(\mathcal {L}_{r_0}\)  the subspace in (\(L1\) ). By (\(L2\) ), \(V(t)\varphi \in \mathcal {L}_{r_0}\)  for \(t\in [0,\tau ]\) . Let \(\chi \in \mathcal {L}_{r_0}^\perp \) . Then \(\langle \chi ,V(t)\varphi \rangle =0\)  holds for \(t\in [0,\tau ]\)  and hence for all \(t\) , see [1]} or [2]}. The argument is as follows. Let without restriction \(H\ge 0\) . Applying the spectral theorem to \(H\) , one has \(F(t):=\langle \chi ,V(t)\varphi \rangle =\int \operatorname{e}^{it\lambda }\operatorname{d}m(\lambda )\)  for a complex measure \(m\)  with \(\operatorname{supp}(m)\subset [0,\infty [\) . Therefore \(F\)  has a continuous extension onto \(\lbrace t\in \Im (t)\ge 0\rbrace \) , which is analytic on \(\lbrace t\in \Im (t)>0\rbrace \) . Moreover, since \(F(t)=0\)  for \(t\in I:=]0,\tau [\) , by Schwarz reflection principle,
there is a further analytic extension of \(F\)  onto \((\mathbb {R}) \cup I\) . This is identical to zero by the principle of analytic continuation. By continuity, \(F=0\) . — It follows \(V(t)\varphi \in \mathcal {L}_{r_0}\)  for all \(t\) .
\(\Box \)
